Pocket companion; containing telegraphic code, tables of standard dimensions of wrought iron pipe, tubes, &c. as manufactured by the National Tube Works Co. and tables of useful information. 1891 - Softcover
Synopsis
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1891 Excerpt: ... FIND THE VELOCITY AND THE QUANTITY DISCHAR(JKI) through a straight smooth, cylindrical iron pipe, whose length is not shorter than 4 times its diameter; knowing its total head, its length, and its diameter, or bore. RULE. Multiply the diameter by the total head; call the product A. Add together the total length of the pipe; and 54 times its diameter. Divide the product A by the sum. Take the square root of the quotient; multiply this square root by the constant number 48. The product will be the first or approximate velocity per second; terms all in feet. For any head, not less than at the rate of 4 feet per mile (about.9 of an inch per 100 ft.), multiply the approximate velocity thus found by the number corresponding to the diameter in feet in the table below. If the pipe is in good order, this last velocity will probably be within 5 to 10 per cent. of the truth. WAT E R--Continued. Then to find the discharge in cubic feet per second, multiply the velocity last found by the area of circular, transverse section of the pipe in square feet. Example: A straight pipe a mile or 5280 feet long, with a diameter of 1 foot has a total fall of 12 feet, measured from the water surface in the reservoir to the center of gravity of its lower end or opening. With what velocity will the water How through it, and how much will be discharged per second? Here the diameter in feet X total head in feet = 1 X 12 = 12 = A. The length in feet is 5280; and 54 times diameter in feet is 54; and these two added together = 5334. And the product A divided by 5334 = j, =.00225. The square root of.00225 is.04743, and.04743 X constant 48 = 2.27 feet per second approximate velocity. The number in above table for 1 foot diameter is also 1; therefore 2.27 X 1 = 2.27 feet per second, the requ...
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